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Paha777 [63]
3 years ago
11

Which graph shows the solution to the system of linear inequalities?

Mathematics
2 answers:
EastWind [94]3 years ago
7 0

Answer:

it's A on e2020

Step-by-step explanation:

i don't know what that other guy is talking about

andrey2020 [161]3 years ago
6 0

Answer:

None of them

Step-by-step explanation:

Both inequalities are shown with < (less than) symbols. Thus the boundary line on both plots of the solution graph should be dashed (not solid) lines. The best choice is the last graph (attached), but the marked line should be dashed, not solid.

The first inequality can be rewritten as ...

... y > (1/4)x -1

The y-intercept of the boundary line will be -1, and the slope of it will be 1/4. Since the inequality symbol is > (not ≥), the boundary line should be dashed. The acceptable values of y are greater than those on the line, so the (blue) shading will be above the line.

The second inequality has a boundary line with a slope of 1 and a y-intercept of +1. Since the inequality symbol is < (not ≤), the boundary line should be dashed (as it is in the attached graph). Acceptable values of y are less than those on the line, so the (red) shading will be below the line.

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Let sin A= -3/5 with 270°&lt; A&lt; 360°. Find the following.<br> sin A/2
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Answer:

1/sqrt10

Step-by-step explanation:

1) Find out cosA using formula (cosA)^2+(sinA)^2=1

The module of cosA= sqrt (1- (-3/5)^2)= sqrt 16/25=4/5

So cosA=-4/5 or cosA=4/5.

Due to the condition 270degrees< A<360 degrees, 0<cosA<1 that's why cosA=4/5.

2) Find sinA/2 using a formula cosA= 1-2sinA/2*sinA/2 where cosA=4/5.

(sinA/2)^2= 0.1

sinA= sqrt 0.1= 1/ sqrt10 or sinA= - sqrt 0.1= -1/sqrt10

But 270°< A< 360°, then 270/2°<A/2<360/2°

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sin A/2= 1/sqrt10

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3 years ago
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Step-by-step explanation:

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